## Dodo1 Group Title Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)2=25(x^2−y^2) at the point (−3,1). The equation of this tangent line can be written in the form y=mx+b where m is: one year ago one year ago

1. Yahoo! Group Title

First Differentiate the function to find the slope.

2. Dodo1 Group Title

thank you but i dont know process. i am not sure about calculation, particulary differentiation

3. Dodo1 Group Title

I mean i know the process

4. zepdrix Group Title

You know the process, but you need assistance with the differentiation portion?

5. Dodo1 Group Title

Yes, when it comes to both chain rule and products. I get really confused

6. zepdrix Group Title

$\large 2(x^2+y^2)^2=25(x^2−y^2)$So we'll be differentiating this implicitly, meaning, as is. No solving for y ahead of time. When we execute the chain rule, a few y' will pop out here and there, and we'll want to solve for that variable. $$\large y'(-3,1)=m$$. This is the m in our equation $$\large y=mx+b$$. Hmm so let's see...

7. zepdrix Group Title

It appears we have no product rule which is nice.

8. Dodo1 Group Title

OK, Y7=(-3,1)=m. right?

9. zepdrix Group Title

We should be able to get a numerical value for m, no x's or y's in our m. I'm not really sure what that Y7 is.

10. Dodo1 Group Title

Ops my mistake

11. zepdrix Group Title

Taking the derivative of the left side gives us, taking the derivative of the outermost function first, the square on the brackets:$\large 2\cdot2(x^2+y^2)^1\color{royalblue}{\frac{d}{dx}(x^2+y^2)}$The blue is due to the chain rule. Understand that part?

12. Dodo1 Group Title

yes, so far good.

13. zepdrix Group Title

$\large 2\cdot2(x^2+y^2)\color{royalblue}{(2x+2yy')}$So we get something like this on the left side.

14. Dodo1 Group Title

i lied, I know product rule, power rule etc, but i dont know what the chain rule is

15. zepdrix Group Title

lol :)

16. zepdrix Group Title

mmmm i always have trouble explaining that one. It's very unintuitive.

17. zepdrix Group Title

hmm

18. Dodo1 Group Title

could you give me example of chain rule problem maybe?

19. zepdrix Group Title

The definition says this,$\large \frac{d}{dx}f\left(g(x)\right) \qquad = \qquad f'(g(x))\cdot g'(x)$When you take the derivative of this, a function within a function, you ~Differentiate the outer function, f in this case. ~Then you make a copy of the inner function, and multiply by it's derivative. Here is a silly example. $\large \frac{d}{dx}\left(x^2\right)^{4}$ In this example we're thinking of $$\large g(x)=x^2$$. And $$\large f(x)=x^4$$. If you remember back to doing compositions of functions, you would understand that when we plug g(x) INTO f(x), our f(x), $$\large f(\color{royalblue}{x})=\left(\color{royalblue}{x}\right)^4$$ will become, $$\large f(\color{royalblue}{g(x)})=\left(\color{royalblue}{g(x)}\right)^4$$ which is $$\large f(\color{royalblue}{x^2})=\left(\color{royalblue}{x^2}\right)^4$$. Applying the chain rule gives us,$\large f'(g(x))=4\left(g(x)\right)^3 g'(x)$See how the rule tells us to make a copy of the inner function and take it's derivative?$\large f'(g(x))=4\left(x^2\right)^3 (x^2)'$$\large f'(g(x))=4\left(x^2\right)^3 (2x)$

20. Dodo1 Group Title

Oo.. I see, Thank you. It was very clear! really cool :)

21. zepdrix Group Title

If you were to simplify this down, you would get,$\large f'(g(x))=8x^7$ Which we know is correct, because we could have easily approached this example without using the chain rule.$\large y=(x^2)^4=x^8$And we know the derivative will match what we got from the chain rule example.

22. zepdrix Group Title

It's a weird concept :d you're like.. making copies of stuff.. but teachers never use that type of language to describe it. So I dunno lol

23. Dodo1 Group Title

no, you explain well. i always get confused and could not tell wether i should use chain or product.

24. zepdrix Group Title

So we applied the derivative of the left side,$\large \frac{d}{dx}(thing)^2 \qquad \rightarrow \qquad 2(thing)(thing)'$ right? :O

25. Dodo1 Group Title

haha thats even more clear haha def, taking note

26. zepdrix Group Title

lol :d

27. Dodo1 Group Title

could we back to the first question?

28. Dodo1 Group Title

i took note

29. zepdrix Group Title

This one? :o $\large 2(x^2+y^2)^2=25(x^2−y^2)$

30. Dodo1 Group Title

yes, haha

31. Dodo1 Group Title

I see the chain rule now, (x^2+y^2)^2?

32. zepdrix Group Title

$\large 2(x^2+y^2)$That's the derivative of the outer function yes? The square on the brackets. But the chain rule is telling us to make a copy of the inside, and take it's derivative.$\large 2(x^2+y^2)(x^2+y^2)'$

33. zepdrix Group Title

$\large 2(x^2+y^2)(2x+2y)$So we should get somethingggggggg like this, yes?

34. Dodo1 Group Title

:O why two in the fron of the bracket? I understand the (2x+2y) tho

35. zepdrix Group Title

You applied the power rule to the outer function $$\large \left(\qquad \right)^2$$

36. Dodo1 Group Title

power rule and chain double? :O

37. zepdrix Group Title

Power rule the outside, make a copy of the inside, take derivative of copy.

38. Dodo1 Group Title

Ok, i understand

39. zepdrix Group Title

Since we took our derivative with respect to x, any time we differentiate a variable that IS NOT x, like y for example, we have to apply the chain rule, and multiply by y'.

40. zepdrix Group Title

$\large 2(x^2+y^2)(2x+2y\color{royalblue}{y'})$See this blue term? We took the derivative of a variable that is not x, namely y, so a y' popped out.

41. Dodo1 Group Title

mmmmmm.... Ok. I think :/

42. Dodo1 Group Title

so whenever its chain rule i put y' in the end?

43. zepdrix Group Title

No. Let's not think of that portion as chain rule, I think that's just going to confuse you. Whenever we take the derivative of y, we will multiply by y'. We don't throw it on the end, we always attach it to the y that we're taking the derivative of.

44. Dodo1 Group Title

right, Ok. i will remember that by my heart.

45. zepdrix Group Title

lol c:

46. Dodo1 Group Title

haha

47. Dodo1 Group Title

whats nxt stp?

48. zepdrix Group Title

Try taking the derivative of the right side, tell me what you get.

49. Dodo1 Group Title

Ok!

50. zepdrix Group Title

Ignore the 25 in front. No chain rule, just take the derivative of the stuff in the brackets.

51. Dodo1 Group Title

25(2x-2y)?

52. zepdrix Group Title

close, but you took the derivative of a y, you need to multiply that part by y'

53. Dodo1 Group Title

25(2x-2yy')?

54. Dodo1 Group Title

its still funny how i have to put y' in the end. is it for implicit derviation?

55. zepdrix Group Title

yes. If we had an equation like this,$\large y=x^2$The derivative gives us,$\large y'=2x$Ok that makes sense.. The y' is telling us what the derivative of y is. But what if we had something like this?$\large y^2=x^2$Taking the derivative,$\large 2y=2x$ Hmm we don't have a derivative term do we?

56. Dodo1 Group Title

Ok, thats cool

57. Dodo1 Group Title

so 2(x^2+y~2)(2x+2y)=25(2x-2yy')

58. zepdrix Group Title

Wasn't there a 2 in front of the equation on the left at the start?? Or was that just a typo?

59. zepdrix Group Title

2(x^2+y^2)^2=25(x^2−y^2)

60. zepdrix Group Title

See that 2 at the front? That would give us an extra factor of 2 in our derivative.

61. Dodo1 Group Title

the question equation is 2(x^2+y^2)^2=25(x^2-y^2)

62. Dodo1 Group Title

so is it 2*2?

63. zepdrix Group Title

yes

64. Dodo1 Group Title

$4(x^2+y^2)(2x+2y)=25(2x-2yy')$

65. zepdrix Group Title

$\large 4(x^2+y^2)(2x+2y\color{royalblue}{y'})=25(2x-2yy')$Woops you missed a y'.

66. Dodo1 Group Title

ops! haha thats derivative of the first equation?

67. zepdrix Group Title

yes. We took the derivative of y on each side, so each y will get a y'

68. Dodo1 Group Title

Ok,

69. zepdrix Group Title

Hmmm from here, I think it's going to be easier to FIRST plug in your coordinate $$\large (-3,1\;)$$ and THEN solve for y'.

70. Dodo1 Group Title

plug both number into the equation?

71. zepdrix Group Title

yes, -3 for all the x's. 1 in for all the y's.

72. zepdrix Group Title

and don't plug anything in for y'. That's what we'll be solving for.